若cosx等于m
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1)sinxcosx=[(sinx+cosx)^2-(sinx^2+cosx^)]/2=(m^2-1)/22)sinx-cosx=正负根号(2-m^2)3)sinx^3+cosx^3=(3m-m^3)
cosx^2=1-sinx^2原式化为:1-sinx^2-2sinxm-3-根号(2m+1)两边同时加2得sinx^2+2sinx+1>m-1-根号(2m+1)即(sinx+1)^2>m-1-根号(2
cos2X=(cosX)^2-(sinX)^2=2*(cosX)^2-1
f(x)=3*根号2*sinx/4*cosx/4+根号6*(cosx/4)的平方-根号6*1/2-m小于或等于0∴(3√2/2)sin(x/2)+(√6/2)[1+cos(x/2)]-√6/2≤m∴(
f(sinx)=cos2x=1-2sin²x∴f(t)=1-2t²∴f(cosx)=1-2cos²x=-cos2x
sin2x-cos4x=2*sinx*cosx-1+8(sinx*cosx)^2由题:2sinx*cosx=m^2-1带进去
sin^2x+cos^2x=1sin^2x+9sin^2x=1sin^2x=1/10Cos2X+Sin2X=cos^2x-sin^2x+2sinxcosx=9sin^2x-sin^2x+6sin^2x
楼上有误.(cosx)^2+2msinx-2m-2-(sinx)^2-1若sinx=1,则2m*0>-2恒成立.否则1-sinx>0,不等式转化成2m>-[(sinx)^2+1]/(1-sinx)①考
f(sinx(π/2-x))=cos(π-2x)f(cosx)=-cos(2x)
f(x)=2cosx^2+2根号3sinxcosx+a=cos2x+根号3sin2x+a+1=2sin(2x+派/6)+a+1a=1x属于[0.pai/2]x=pai/6取最大值f(x)=4x=pia
f(x)=2(cosx)^2+根号3sin2x+m=cos2x+根号3sin2x+m+2化简:f(x)=2sin(2x+30度)+m+2因为f(x)在大于等于0小于等于派上,所以(2x+30度)在大于
secx-1=(1/cosx)-1=(1/cosx)-(cosx/cosx)=(1-cosx)/cosx
答:f(x)=√3cosx/4sinx/4+cos²x/4=√3/2sinx/2+1/2cosx/2+1/2=sin(x/2+π/6)+1/2(1).当f(x)=1时,sin(x/2+π/6
f'(x)=-sinx
很简单,换元就搞定,sinx+cosx设为t,则sinxcosx=(t的平方-1)/2,然后就可以换成关于t的二次函数,当然t属于(-√2√2),把m移到右边去,算出左边换成关于t的二次函数的最小值,
sinx+cosx=m平方一下得sin^2x+2sinxcosx+cos^2x=m^2sinxcosx=(m^2-1)/2sin^3x+cos^3x=(sinx+cosx)(sin^2-sinxcos
f(sinx)=1+1-cos2x=1+2sin²xf(cosx)=1+2cos²x=cos2x+2再问:1-cos2x=???再答:cos2x=cos²x-sin&su
等于+-根号下(1+cos)/2
[(1+sinx)/cosx]/[cosx/(sinx-1)]=[(sinx)^2-1]/(cosx)^2=-(cosx)^2/(cosx)^2=-1因为(1+sinx)/cosx=-0.5所以cos
[2π/3+2kπ,4π/3+2kπ]k为Z.